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Question 1204569: Explain why, for any values of a, b, and c, the equations
2x + 2y + 5z = a
−3x + y − 2z = b
x + z = c
always have a unique solution.
Find this unique solution (in terms of a, b, and c)
Answer by math_tutor2020(3820)   (Show Source):
You can put this solution on YOUR website!
The given system
converts to this matrix
Normally matrices do not have separating lines, but I think it's beneficial to have them.
I'll place the items in a table like this
Let's apply row operations to zero out the entries below each pivot.
| 1 | 1 | 5/2 | a/2 | (1/2)*R1 --> R1 | | -3 | 1 | -2 | b | | | 1 | 0 | 1 | c | |
Notation like (1/2)*R1 --> R1 means we take half of each element in row 1 (aka R1). Then store the results in R1.
| 1 | 1 | 5/2 | a/2 | | | 0 | 4 | 11/2 | 3a/2+b | R2 + 3*R1 --> R2 | | 1 | 0 | 1 | c | |
Something like R2 + 3*R1 --> R2 will mean that we triple everything in R1, then add those results to R2. Store the results in R2.
| 1 | 1 | 5/2 | a/2 | | | 0 | 4 | 11/2 | 3a/2+b | | | 0 | -1 | -3/2 | c - a/2 | R3 - R1 --> R3 |
| 1 | 1 | 5/2 | a/2 | | | 0 | 1 | 11/8 | 3a/8+b/4 | (1/4)*R2 --> R2 | | 0 | -1 | -3/2 | c - a/2 | |
| 1 | 1 | 5/2 | a/2 | | | 0 | 1 | 11/8 | 3a/8+b/4 | | | 0 | 0 | -1/8 | -a/8 + b/4 + c | R3 + R2 --> R3 |
| 1 | 1 | 5/2 | a/2 | | | 0 | 1 | 11/8 | 3a/8+b/4 | | | 0 | 0 | 1 | a-2b-8c | -8*R3 --> R3 |
At this point the matrix is in row echelon form (REF), but we haven't reached RREF just yet.
Let's keep row reducing until all of the non-pivot entries are turned to 0.
| 1 | 1 | 5/2 | a/2 | | | 0 | 1 | 0 | -a+3b+11c | R2 - (11/8)*R3 --> R2 | | 0 | 0 | 1 | a-2b-8c | |
| 1 | 1 | 0 | -2a+5b+20c | R1 - (5/2)*R3 --> R3 | | 0 | 1 | 0 | -a+3b+11c | | | 0 | 0 | 1 | a-2b-8c | |
| 1 | 0 | 0 | -a+2b+9c | R1 - R2 --> R1 | | 0 | 1 | 0 | -a+3b+11c | | | 0 | 0 | 1 | a-2b-8c | |
The matrix is now in Reduced Row Echelon Form (RREF).
The system has exactly one solution, and it is when:
x = -a+2b+9c
y = -a+3b+11c
z = a-2b-8c
Confirmation using WolframAlpha
https://www.wolframalpha.com/input/?i=2x%2B2y%2B5z%3Da%2C-3x%2By-2z%3Db%2Cx%2Bz%3Dc
The search input to type in is "2x+2y+5z=a,-3x+y-2z=b,x+z=c" without quotes.
The CAS feature in GeoGebra can also be used.
There are two options to input
Either
Solve[{2x+2y+5z=a,-3x+y-2z=b,x+z=c}]
or
ReducedRowEchelonForm[{{2, 2, 5, a}, {-3, 1, -2, b}, {1, 0, 1, c}}]
The way GeoGebra handles matrices is that they are a collection of lists.
Each list is enclosed in curly braces, which represents a particular row.
There are likely other CAS based matrix calculators out there that can offer similar features.
Some more practice with matrix row reduction
https://www.algebra.com/algebra/college/linear/Linear_Algebra.faq.question.1204100.html
and
https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Matrices-and-determiminant.faq.question.1203997.html
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